MHT CET · Physics · Current Electricity
In a Wheatstone bridge, three resistance \(P, Q\) and \(R\) are connected in the three arms and the fourth arm is formed by two resistances \(S_1\) and \(S_2\) connected in parallel. The condition for the bridge to be balanced is
- A \(\frac{P}{Q}=\frac{2 R}{S_1+S_2}\)
- B \(\frac{P}{Q}=\frac{R\left(S_1+S_2\right)}{2 S_1 S_2}\)
- C \(\frac{P}{Q}=\frac{R\left(S_1+S_2\right)}{S_1 S_2}\)
- D \(\frac{P}{Q}=\frac{R\left(S_1 S_2\right)}{S_1+S_2}\)
Answer & Solution
Correct Answer
(C) \(\frac{P}{Q}=\frac{R\left(S_1+S_2\right)}{S_1 S_2}\)
Step-by-step Solution
Detailed explanation
For balanced Wheatstone's bridge we have
\(
\frac{P}{Q}=\frac{R}{S}
\)
Where \(S\) is the equivalent resistance of \(S_1\) and \(S_2\) in parallel
\(
S=\frac{S_1 S_2}{S_1+S_2}
\)
\(
\therefore \frac{P}{Q}=\frac{R\left(S_1+S_2\right)}{S_1 S_2}
\)
\(
\frac{P}{Q}=\frac{R}{S}
\)
Where \(S\) is the equivalent resistance of \(S_1\) and \(S_2\) in parallel
\(
S=\frac{S_1 S_2}{S_1+S_2}
\)
\(
\therefore \frac{P}{Q}=\frac{R\left(S_1+S_2\right)}{S_1 S_2}
\)
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